There are two different positive three-digit integers that are exactly forty-eight times the sum of their digits. Find both three-digit positive integers that satisfy this condition.Source: mathcontest.olemiss.edu 10/27/2008

SOLUTIONThis problem is similar to the problem titled “11 Times” dated 12/8/2008.

We will list a few multiples of 48 starting from 48, 96, 144, etc. Then, we will go through each of them and determine if the number is equal to 48 times the sum of its digits.

Example 148 is not a solution.

Example 296 is not a solution.

Example 3144 is not a solution.

The table below displays the operations:

48 × 1 = 48

4 + 8 ≠ 1

48 × 7 = 336

3 + 3 + 6 ≠ 7

48 × 13 = 624

6 + 2 + 4 ≠ 13

48 × 2 = 96

9 + 6 ≠ 2

48 × 8 = 384

3 + 8 + 4 ≠ 8

48 × 14 = 672

6 + 7 + 2 ≠ 14

48 × 3 = 144

1 + 4 + 4 ≠ 3

48 × 9 = 432

4 + 3 + 2 = 9

48 × 15 = 720

7 + 2 + 0 ≠ 15

48 × 4 = 192

1 + 9 + 2 ≠ 4

48 × 10 = 480

4 + 8 + 0 ≠ 10

48 × 16 = 768

7 + 6 + 8 ≠ 16

48 × 5 = 240

2 + 4 + 0 ≠ 5

48 × 11 = 528

5 + 2 + 8 ≠11

48 × 17 = 816

8 + 1 + 6 ≠ 17

48 × 6 = 288

2 + 8 + 8 ≠ 6

48 × 12 = 576

5 + 7 + 6 ≠ 12

48 × 18 = 864

8 + 6 + 4 = 18

Thus, 432 is 48 times the sum of its digits and 864 is 48 times the sum of its digits.